Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi everyone,

Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the completed free Lie algebra on the same number of generators as $G$, by the (formal) logarithm of the defining relations of $G$ (see edit below).

$G$ is called 1-formal if its Malcev Lie algebra is isomorphic as a filtered Lie algebra to the degree completion of a finitely presented Lie algebra with quadratic relations. (If $X$ is a path-connected topological space such that $G=\pi_1(X)$, this is equivalent to say that the Malcev Lie algebra of $G$ is isomorphic to the holonomy Lie algebra of $X$)

It is quite well known that fundamental groups of complementary of hyperplane arrangements in $\mathbb{C}^n$ (e.g. pure braid groups) are 1-formal.

On the one hand, an important fact in the study of these groups is that they are (under some conditions on the underlying arrangement) iterated "almost-direct product" of free groups, and free groups are themselves 1-formal. An almost direct product is a semi-direct product $H\rtimes K$ for which the action of $K$ on the abelianization of $H$ is trivial. On the other hand, according to this interesting survey http://www.arxiv.com/abs/0903.2307 the direct product of two 1-formal groups is again 1-formal, even if no proof is given (it's probably easy..)

So it is temptating to ask:

is an almost-direct product of 1-formal groups again 1-formal ?

Edit: Some details about the construction and its relation with almost direct product. If $G$ is an abelian group, then one can take the tensor product $G \otimes_{\mathbb{Z}} \mathbb{Q}$ leading to a uniquely divisible abelian group. The Malcev construction extends this to any nilpotent group, leading to a uniquely divisible nilpotent group. Let $G^{(0)}=G$, $G^{(n+1)}=[G^{(n)},G]$ be the lower central serie, then the quotients $G/G^{(n)}$ are nilpotent by construction.

The Malcev completion of $G$ is the inverse limit of the $(G/G^{(n)}) \otimes_{\mathbb{Z}} \mathbb{Q}$. It is a pro-unipotent group, hence it has a (pro-nilpotent) Lie algebra $\mathfrak g$. On the other hand, one can define a Lie algebra by $$gr\ G=\mathbb{Q}\otimes_{\mathbb{Z}} \bigoplus G^{(n+1)}/G^{(n)}$$ the bracket being induced by the commutator in $G$. It as a graded Lie algebra, and in fact it is the associated graded of the filtered Lie algebra $\mathfrak g$.

A group is called 1-formal if there is an isomorphism of filtered Lie algebra $gr\ G \cong \mathfrak g$. Now, the point is that almost-direct product behave well with respect to the lower central serie. If $G=G_1\rtimes G_2$ is an almost-direct product, then it seems to me that a result of Falk and Randell implies that we have $$G^{(n)}=G_1^{(n)}\rtimes G_2^{(n)}$$ and $$gr\ G = gr\ G_1 \rtimes gr\ G_2$$.

share|improve this question
add comment

1 Answer

up vote 11 down vote accepted

Yes, the direct product of two 1-formal groups is again 1-formal, and so is the free product of two 1-formal groups. A proof is given in arxiv:0902.1250, Proposition 9.2.

And no, the almost direct product of two 1-formal groups need not be 1-formal. A proof is given in the same paper, Example 8.2. The group in question is a semi-direct product of the form $G=F_4\rtimes F_1$, where $F_n$ is the free group of rank $n$, which is of course 1-formal. The action of $F_1$ on $F_4$ is given by a certain pure braid $\beta \in P_4$, acting via the Artin representation on $F_4$; thus, the action is trivial on $H_1(F_4)$. For this extension, the ``tangent cone formula" fails: the tangent cone to the characteristic variety $V_2(G)$ is strictly included in the resonance variety $R_2(G)$. In view of Theorem A from the cited paper, the group $G$ is not 1-formal.

It is worth noting that $G$ is the fundamental group of the complement of a certain link of 5 great circles in $S^3$. Alternatively, $G$ can be realized as the fundamental group of the complement of an arrangement of 5 planes in $\mathbb{R}^4$, meeting transversely at the origin (of course, this real arrangement cannot be isotoped to an arrangement of 5 complex lines in $\mathbb{C}^2$). For more details on the construction and properties of such arrangements, see arxiv:math.GT/9712251. In particular, the pure braid $\beta$ is described there in Propositions 4.4, 4.6, and 4.9.

share|improve this answer
    
Thanks ! I didn't really believe that it was true in full generality, but I still thought that it was true in resaonable situations like... well the situation you describe, so this is a very interesting counterexample.. Are there still some specific techniques to prove that a given almost-direct product of 1-formal group is 1 formal ? –  Adrien Sep 16 '11 at 6:28
    
There are not many general techniques: it all depends on what the automorphisms defining the almost-direct products are. Intuitively, I think, the deeper into the Johnson filtration your automorphism is (i.e., the more it looks like the identity through the eye of the successive nilpotent quotients), the more likely it is that the extension will be non-1-formal. –  Alex Suciu Sep 16 '11 at 14:13
    
For instance, it is known that the full McCool group $P\Sigma_n$ (the group of basis-conjugating automorphisms of $F_n$) is 1-formal; see <a href="dx.doi.org/10.1142/S0218216509007257"; rel="nofollow">this</a> paper by Berceanu and Papadima. But, as far as I know, it is an open question whether the upper-triangular McCool group $P\Sigma_n^+$ is 1-formal (for $n\ge 4$). Note that this group can be realized as an iterated almost-direct product of free groups, $P\Sigma_n^+=F_{n-1}\rtimes \cdots \rtimes F_2\rtimes F_1$, much the same as the pure braid group $P_n$. –  Alex Suciu Sep 16 '11 at 18:17
    
Thanks, actually I'm interested in surface pure braid groups. I know Bezrukavnikovproved they are 1-formal, but it seems to me that they are iterated alomost direct extension of the fundamental group of the surface by free groups, and I'm wondering if this fact can be exploited to prove the 1-formality. –  Adrien Sep 17 '11 at 10:26
1  
Careful: the surface pure braid groups $P_{g,n}$ are always 1-formal, except when $g=1$ and $n\ge 3$, when they are not. A proof that $P_{1,n}$ is not 1-formal for $n\ge 3$ is given in arxiv.org/abs/0902.1250, Example 10.1, again using the ``tangent cone formula". –  Alex Suciu Sep 18 '11 at 2:08
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.